Question

the graph of the line $4x + 3y = -6$ is shown. which three ordered pairs are in the solution set of the inequality $4x + 3y \leq -6$?
a $(2, -1)$
b $(1, -3)$
c $(0, -4)$
d $(-1, -3)$

Explanation:

Step1: Test point A (2, -1)

Substitute $x=2, y=-1$ into $4x + 3y$:
$4(2) + 3(-1) = 8 - 3 = 5$
Check if $5 \leq -6$: No.

Step2: Test point B (1, -3)

Substitute $x=1, y=-3$ into $4x + 3y$:
$4(1) + 3(-3) = 4 - 9 = -5$
Check if $-5 \leq -6$: No.

Step3: Test point C (0, -4)

Substitute $x=0, y=-4$ into $4x + 3y$:
$4(0) + 3(-4) = 0 - 12 = -12$
Check if $-12 \leq -6$: Yes.

Step4: Test point D (-3, -2)

Substitute $x=-3, y=-2$ into $4x + 3y$:
$4(-3) + 3(-2) = -12 - 6 = -18$
Check if $-18 \leq -6$: Yes.

Step5: Identify valid points (note: assuming a typo, and adding the line's intercept (-3, 2) as a valid solution since it lies on $4x+3y=-6$, which satisfies $\leq$)

For $(-3,2)$: $4(-3)+3(2)=-12+6=-6$, so $-6 \leq -6$: Yes.

Answer:

C. (0, -4), D. (-3, -2), and (-3, 2) (assuming a missing option; if only given options, C and D are valid, with the third being a point on the line like (-3,2))

Note: If we only use the provided visible options, the valid ones are C and D. To meet the "THREE" requirement, a point on the line such as $(-3, 2)$ (which satisfies $4x+3y=-6$) is also part of the solution set.